Power of a point

Orthogonal circle (green)

For any point ${\displaystyle P}$ outside of the circle ${\displaystyle c}$ there are two tangent points ${\displaystyle T_{1},T_{2}}$ on circle ${\displaystyle c}$, which have equal distance to ${\displaystyle P}$. Hence the circle ${\displaystyle o}$ with center ${\displaystyle P}$ through ${\displaystyle T_{1}}$ passes ${\displaystyle T_{2}}$, too, and intersects ${\displaystyle c}$ orthogonal:

• The circle with center ${\displaystyle P}$ and radius ${\displaystyle {\sqrt {\Pi (P)}}}$ intersects circle ${\displaystyle c}$ orthogonal.

Angle between two circles

If the radius ${\displaystyle \rho }$ of the circle centered at ${\displaystyle P}$ is different from ${\displaystyle {\sqrt {\Pi (P)}}}$ one gets the angle of intersection ${\displaystyle \varphi }$ between the two circles applying the Law of cosines (see the diagram):

${\displaystyle \rho ^{2}+r^{2}-2\rho r\cos \varphi =|PO|^{2}}$

${\displaystyle \rightarrow \ \cos \varphi ={\frac {\rho ^{2}+r^{2}-|PO|^{2}}{2\rho r}}={\frac {\rho ^{2}-\Pi (P)}{2\rho r}}}$

(${\displaystyle PS_{1}}$ and ${\displaystyle OS_{1}}$ are normals to the circle tangents.)

If ${\displaystyle P}$ lies inside the blue circle, then ${\displaystyle \Pi (P)<0}$ and ${\displaystyle \varphi }$ is always different from ${\displaystyle 90^{\circ }}$.

If the angle ${\displaystyle \varphi }$ is given, then one gets the radius ${\displaystyle \rho }$ by solving the quadratic equation

${\displaystyle \rho ^{2}-2\rho r\cos \varphi -\Pi (P)=0}$.

Secant-, chord-theorem

For the intersecting secants theorem and chord theorem the power of a point plays the role of an invariant:

• Intersecting secants theorem: For a point ${\displaystyle P}$ outside a circle ${\displaystyle c}$ and the intersection points ${\displaystyle S_{1},S_{2}}$ of a secant line ${\displaystyle g}$ with ${\displaystyle c}$ the following statement is true: ${\displaystyle |PS_{1}|\cdot |PS_{2}|=\Pi (P)}$, hence the product is independent of line ${\displaystyle g}$. If ${\displaystyle g}$ is tangent then ${\displaystyle S_{1}=S_{2}}$ and the statement is the tangent-secant theorem.
• Intersecting chords theorem: For a point ${\displaystyle P}$ inside a circle ${\displaystyle c}$ and the intersection points ${\displaystyle S_{1},S_{2}}$ of a secant line ${\displaystyle g}$ with ${\displaystyle c}$ the following statement is true: ${\displaystyle |PS_{1}|\cdot |PS_{2}|=-\Pi (P)}$, hence the product is independent of line ${\displaystyle g}$.

Let ${\displaystyle P}$ be a point and ${\displaystyle c_{1},c_{2}}$ two non concentric circles with centers ${\displaystyle O_{1},O_{2}}$ and radii ${\displaystyle r_{1},r_{2}}$. Point ${\displaystyle P}$ has the power ${\displaystyle \Pi _{i}(P)}$ with respect to circle ${\displaystyle c_{i}}$. The set of all points ${\displaystyle P}$ with ${\displaystyle \Pi _{1}(P)=\Pi _{2}(P)}$ is a line called radical axis. It contains possible common points of the circles and is perpendicular to line ${\displaystyle {\overline {O_{1}O_{2}}}}$.

Similarity points are an essential tool for Steiner's investigations on circles.[5]

Given two circles

${\displaystyle \ c_{1}:({\vec {x}}-{\vec {m}}_{1})-r_{1}^{2}=0,\quad c_{2}:({\vec {x}}-{\vec {m}}_{2})-r_{2}^{2}=0\ .}$

A homothety (similarity) ${\displaystyle \sigma }$, that maps ${\displaystyle c_{1}}$ onto ${\displaystyle c_{2}}$ stretches (jolts) radius ${\displaystyle r_{1}}$ to ${\displaystyle r_{2}}$ and has its center ${\displaystyle Z:{\vec {z}}}$ on the line ${\displaystyle {\overline {M_{1}M_{2}}}}$, because ${\displaystyle \sigma (M_{1})=M_{2}}$. If center ${\displaystyle Z}$ is between ${\displaystyle M_{1},M_{2}}$ the scale factor is ${\displaystyle s=-{\tfrac {r_{2}}{r_{1}}}}$. In the other case ${\displaystyle s={\tfrac {r_{2}}{r_{1}}}}$. In any case:

${\displaystyle \sigma ({\vec {m}}_{1})={\vec {z}}+s({\vec {m}}_{1}-{\vec {z}})={\vec {m}}_{2}}$.

Inserting ${\displaystyle s=\pm {\tfrac {r_{2}}{r_{1}}}}$ and solving for ${\displaystyle {\vec {z}}}$ yields:

${\displaystyle {\vec {z}}={\frac {r_{1}{\vec {m}}_{2}\mp r_{2}{\vec {m}}_{1}}{r_{1}\mp r_{2}}}}$.

Similarity points of two circles: various cases

Point

$${\displaystyle E:{\vec {e}}={\frac {r_{1}{\vec {m}}_{2}-r_{2}{\vec {m}}_{1}}{r_{1}-r_{2}}}}$$

is called the exterior similarity point and

$${\displaystyle I:{\vec {i}}={\frac {r_{1}{\vec {m}}_{2}+r_{2}{\vec {m}}_{1}}{r_{1}+r_{2}}}}$$

is called the inner similarity point.

In case of ${\displaystyle M_{1}=M_{2}}$ one gets ${\displaystyle E=I=M_{i}}$.
In case of ${\displaystyle r_{1}=r_{2}}$: ${\displaystyle E}$ is the point at infinity of line ${\displaystyle {\overline {M_{1}M_{2}}}}$ and ${\displaystyle I}$ is the center of ${\displaystyle M_{1},M_{2}}$.
In case of ${\displaystyle r_{1}=|EM_{1}|}$ the circles touch each other at point ${\displaystyle E}$ inside (both circles on the same side of the common tangent line).
In case of ${\displaystyle r_{1}=|IM_{1}|}$ the circles touch each other at point ${\displaystyle I}$ outside (both circles on different sides of the common tangent line).

Further more:

• If the circles lie disjoint (the discs have no points in common), the outside common tangents meet at ${\displaystyle E}$ and the inner ones at ${\displaystyle I}$.
• If one circle is contained within the other, the points ${\displaystyle E,I}$ lie within both circles.
• The pairs ${\displaystyle M_{1},M_{2};E,I}$ are projective harmonic conjugate: Their cross ratio is ${\displaystyle (M_{1},M_{2};E,I)=-1}$.

Monge's theorem states: The outer similarity points of three disjoint circles lie on a line.

Similarity points of two circles and their common power

Let ${\displaystyle c_{1},c_{2}}$ be two circles, ${\displaystyle E}$ their outer similarity point and ${\displaystyle g}$ a line through ${\displaystyle E}$, which meets the two circles at four points ${\displaystyle G_{1},H_{1},G_{2},H_{2}}$. From the defining property of point ${\displaystyle E}$ one gets

${\displaystyle {\frac {|EG_{1}|}{|EG_{2}|}}={\frac {r_{1}}{r_{2}}}={\frac {|EH_{1}|}{|EH_{2}|}}\ }$

${\displaystyle \rightarrow \ |EG_{1}|\cdot |EH_{2}|=|EH_{1}|\cdot |EG_{2}|\ }$

and from the secant theorem (see above) the two equations

${\displaystyle |EG_{1}|\cdot |EH_{1}|=\Pi _{1}(E),\quad |EG_{2}|\cdot |EH_{2}|=\Pi _{2}(E).}$

Combining these three equations yields:

$${\displaystyle {\begin{aligned}\Pi _{1}(E)\cdot \Pi _{2}(E)&=|EG_{1}|\cdot |EH_{1}|\cdot |EG_{2}|\cdot |EH_{2}|\\&=|EG_{1}|^{2}\cdot |EH_{2}|^{2}=|EG_{2}|^{2}\cdot |EH_{1}|^{2}\ .\end{aligned}}}$$

Hence:

$${\displaystyle |EG_{1}|\cdot |EH_{2}|=|EG_{2}|\cdot |EH_{1}|={\sqrt {\Pi _{1}(E)\cdot \Pi _{2}(E)}}}$$

(independent of line ${\displaystyle g}$ !).

The analog statement for the inner similarity point ${\displaystyle I}$ is true, too.

The invariants ${\textstyle {\sqrt {\Pi _{1}(E)\cdot \Pi _{2}(E)}},\ {\sqrt {\Pi _{1}(I)\cdot \Pi _{2}(I)}}}$ are called by Steiner common power of the two circles (gemeinschaftliche Potenz der beiden Kreise bezüglich ihrer Ähnlichkeitspunkte).[6]

The pairs ${\displaystyle G_{1},H_{2}}$ and ${\displaystyle H_{1},G_{2}}$ of points are antihomologous points. The pairs ${\displaystyle G_{1},G_{2}}$ and ${\displaystyle H_{1},H_{2}}$ are homologous.[7][8]

Common power of two circles: application

Circles tangent to two circles

For a second secant through ${\displaystyle E}$:

${\displaystyle |EH_{1}|\cdot |EG_{2}|=|EH'_{1}|\cdot |EG'_{2}|}$

From the secant theorem one gets:

The four points ${\displaystyle H_{1},G_{2},H'_{1},G'_{2}}$ lie on a circle.

And analogously:

The four points ${\displaystyle G_{1},H_{2},G'_{1},H'_{2}}$ lie on a circle, too.

Because the radical lines of three circles meet at the radical (see: article radical line), one gets:

The secants ${\displaystyle {\overline {H_{1}H'_{1}}},\;{\overline {G_{2}G'_{2}}}}$ meet on the radical axis of the given two circles.

Moving the lower secant (see diagram) towards the upper one, the red circle becomes a circle, that is tangent to both given circles. The center of the tangent circle is the intercept of the lines ${\displaystyle {\overline {M_{1}H_{1}}},{\overline {M_{2}G_{2}}}}$. The secants ${\displaystyle {\overline {H_{1}H'_{1}}},{\overline {G_{2}G'_{2}}}}$ become tangents at the points ${\displaystyle H_{1},G_{2}}$. The tangents intercept at the radical line ${\displaystyle p}$ (in the diagram yellow).

Similar considerations generate the second tangent circle, that meets the given circles at the points ${\displaystyle G_{1},H_{2}}$ (see diagram).

All tangent circles to the given circles can be found by varying line ${\displaystyle g}$.

Positions of the centers

Circles tangent to two circles

If ${\displaystyle X}$ is the center and ${\displaystyle \rho }$ the radius of the circle, that is tangent to the given circles at the points ${\displaystyle H_{1},G_{2}}$, then:

${\displaystyle \rho =|XM_{1}|-r_{1}=|XM_{2}|-r_{2}}$

${\displaystyle \rightarrow \ |XM_{2}|-|XM_{1}|=r_{2}-r_{1}.}$

Hence: the centers lie on a hyperbola with

foci ${\displaystyle M_{1},M_{2}}$,

distance of the vertices ${\displaystyle 2a=r_{2}-r_{1}}$,

center ${\displaystyle M}$ is the center of ${\displaystyle M_{1},M_{2}}$ ,

linear eccentricity ${\displaystyle c={\tfrac {|M_{1}M_{2}|}{2}}}$ and

${\displaystyle \ b^{2}=e^{2}-a^{2}={\tfrac {|M_{1}M_{2}|^{2}-(r_{2}-r_{1})^{2}}{4}}}$.

Considerations on the outside tangent circles lead to the analog result:

If ${\displaystyle X}$ is the center and ${\displaystyle \rho }$ the radius of the circle, that is tangent to the given circles at the points ${\displaystyle G_{1},H_{2}}$, then:

${\displaystyle \rho =|XM_{1}|+r_{1}=|XM_{2}|+r_{2}}$

${\displaystyle \rightarrow \ |XM_{2}|-|XM_{1}|=-(r_{2}-r_{1}).}$

The centers lie on the same hyperbola, but on the right branch.

See also Problem of Apollonius.

Power of a point with respect to a sphere